a)ab(a+b)-bc(b+c)+ac(a-c)

=ab(a+b)-bc(b+c)+ac\([\left(a+b\right)-\left(b+c\right)]\)

=ab(a+b)-bc(b+c)+ac(a+b)-ac(b+c)

=(a+b)(ab+ac)-(b+c)(bc+ac)

=(a+b)a(b+c)-(b+c)c(b+a)

=(a+b)(b+c)(a-c)

\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(=a^2b+abc+ca^2+ab^2+b^2c+abc+abc+bc^2+ac^2-abc\)

\(=a^2b+a^2c+ab^2+b^2c+c^2a+bc^2+ac^2+2abc\)

\(=\left(a^2b+ba^2+abc\right)+\left(b^2c+c^2b+abc\right)+\left(ac^2+ca^2\right)\)

\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+c\right)\)

\(=\left(a+b+c\right)\left(ab+bc\right)+ac\left(a+c\right)\)

\(=b.\left(a+b+c\right)\left(a+c\right)+ac\left(a+c\right)\)

\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)

\(=\left(a+c\right)\left[b.\left(a+b\right)+c.\left(a+b\right)\right]\)

\(=\left(a+c\right)\left(a+b\right)\left(b+c\right)\)

https://h7.net/hoi-dap/toan-8/phan-h-da-thuc-abc-ab-bc-ca-a-b-c-1-thanh-nhan-tu-faq382483.html

nhân hả bạn

(a + b + c)(ab + bc + ac) - abc

= a²b + abc +a²c + ab² + b²c + abc + abc + bc² + ac²

= (a²b + 2abc + bc²) + (ac² + a²c) + (ab² + b²c)

= b(a² + 2ac + c²) + ac(c + a) + b²(a + c)

= b(a + c) + ac(a + c) + b²(a + c)

= (a + c)[b(a + c) + ac + b²]

= (a + c)(ab + bc + ac + b²)

= (a + c)[b(a + b) + c(a + b)]

= (a + c)(b + c)(a + b)

\(\left(a+b+c\right)\left(ab+bc+ac\right)-abc\)

\(=a^2b+abc+a^2c+b^2a+b^2c+abc+abc+c^2b+c^2a-abc\)

\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a^2+b^2+2ab-2ab\right)+2abc\)

\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2-2abc+2abc\)

\(=\left(a+b\right)\left(ab+c^2+ca+cb\right)\)

\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

abc - (ab + ac + bc) + (a + b + c) - 1

= abc - bc - ab + b - ac + c + a - 1

= bc(a - 1) - b(a - 1) - c(a - 1) + (a - 1)

= (a - 1)(bc - b - c + 1)

= (a - 1)(b - 1)(c - 1)

bạn giữ nguyên ab(a+b),mấy cái kia triển khai ra,xong rồi nhóm hạng tử thu gọn lại,dc nhân tử chung là (a+b),xong rồi đưa (a+b) lm nhân tử chung,bên trong thu gọn từ từ

d) (b+c)(b+a)(c-a)

c) (b-1)(ac+1-a-c)

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